Scaling a math function from a specific vertex
I'm not sure what the best way to scale that with Blender would be, but I'd look at just adjusting your equation. Here's your $y$ equation:

$$y=\begin{cases}
0.5 &amp; \text{if } u &lt; 0 \\
\frac{u^2}{u^2+(1-u)^2}+0.5 &amp; \text{if } 0 \leq u \leq 1\\
1.5 &amp; \text{if } u &gt; 1
\end{cases}$$

Notice that the main part of the middle equation, $\frac{u^2}{u^2+(1-u)^2}$, normally has a range of $[0,1]$ on the interval $[0,1]$. You are offsetting it by $0.5$. If we scale it, and change the offset of the third piece, we can place it wherever. Use this equation instead:

$$y=\begin{cases}
0.5 &amp; \text{if } u &lt; 0 \\
k \cdot \frac{\left(\frac{u}{k}\right)^2}{\left(\frac{u}{k}\right)^2+\left(1-\frac{u}{k}\right)^2}+0.5 &amp; \text{if } 0 \leq \frac{u}{k} \leq 1\\
k + 0.5 &amp; \text{if } \frac{u}{k} &gt; 1
\end{cases}$$

where $k$ is your scalar. You can try it online using Desmos here. (Thanks to Rich Sedman for help with scaling horizontally).

In a Python expression, it would be this (substituting in whatever value for k):

(0.5 if u&lt;0 else (k*((u/k)**2/((u/k)**2+(1-(u/k))**2)+0.5) if (u/k)&lt;=1 else k+0.5))*v